# Vector field of a flow's cotangent lift is Hamiltonian

Let $M$ be a smooth manifold, and $X\in \mathfrak{X}(M)$ be a complete vector field. Then for all $t \in \Bbb R$ we have the flow $\Phi_{t,X}\colon M \to M$ of $X$. We can consider the cotangent lift $\widehat{\Phi_{t,X}}\colon T^*M \to T^*M$ of each flow time$^1$. By properties of the flow, we have that $\widehat{\Phi_{t,X}} = \Phi_{t,\hat{X}}$ for some complete vector field $\hat{X} \in\mathfrak{X}(T^*M)$.

We consider in $T^*M$ the standard symplectic strutcure $\omega_{\rm can}$. I want to check that $\hat{X}$ is Hamiltonian, and find a Hamiltonian function $H\colon T^*M \to \Bbb R$ (i.e., such that $\omega_{\rm can}(\hat{X},\cdot) = {\rm d}H$).

Since cotangent lifts are symplectomorphisms, I know that $\hat{X}$ is symplectic, and so locally Hamiltonian. Also: $$\pi\circ \Phi_{t,\hat{X}} = \Phi_{t,X}\circ \pi \implies {\rm d}\pi_\xi(\hat{X}_\xi) = X_{\pi(\xi)}, \qquad \mbox{for all }\xi \in T^*M.$$The only map $T^*M \to \Bbb R$ I can immediately think of is $\alpha(\hat{X})$, where $\alpha$ is the tautological $1$-form, but this doesn't seem to work and I don't know what else to do here. Maybe treat the situation with brute force in coordinates? Please help.

If $f\colon M\to N$ is a diffeomorphism, the cotangent lift is $\widehat{f}\colon T^*M \to T^*N$, given by $\widehat{f}(\xi) = \xi \circ ({\rm d}f_{\pi(\xi)})^{-1}$, where $\pi\colon T^*M \to M$ is the canonical projection.

We have that $\omega_{can}=-d\alpha$. So $$\iota_{\hat{X}}\omega_{can}=\iota_{\hat{X}} (-d\alpha)=-\iota_{\hat{X}}d\alpha=-\mathcal{L}_{\hat{X}}\alpha+d\iota_{\hat{X}}\alpha=d\iota_{\hat{X}}\alpha=d(\alpha(\hat{X})),$$ where we have used that $\mathcal{L}_{\hat{X}}\alpha=0$, which is true since $\widehat{X}$ yields the lift of the flow.